NONEUCLIDEAN GEOMETRIES


 Nicholas Houston
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1 Chapter 3 NONEUCLIDEN GEOMETRIES In the previous chapter we began by adding Euclid s Fifth Postulate to his five common notions and first four postulates. This produced the familiar geometry of the Euclidean plane in which there exists precisely one line through a given point parallel to a given line not containing that point. In particular, the sum of the interior angles of any triangle was always 180 no matter the size or shape of the triangle. In this chapter we shall study various geometries in which parallel lines need not exist, or where there might be more than one line through a given point parallel to a given line not containing that point. For such geometries the sum of the interior angles of a triangle is then always greater than 180 or always less than 180. This in turn is reflected in the area of a triangle which turns out to be proportional to the difference between 180 and the sum of the interior angles. First we need to specify what we mean by a geometry. This is the idea of an bstract Geometry introduced in Section 3.1 along with several very important examples based on the notion of projective geometries, which first arose in Renaissance art in attempts to represent threedimensional scenes on a twodimensional canvas. Both Euclidean and hyperbolic geometry can be realized in this way, as later sections will show. 3.1 BSTRCT ND LINE GEOMETRIES. One of the weaknesses of Euclid s development of plane geometry was his definition of points and lines. He defined a point as... that which has no part and a line as... breadthless length. These really don t make much sense, yet for over 2,000 years everything he built on these definitions has been regarded as one of the great achievements in mathematical and intellectual history! Because Euclid s definitions are not very satisfactory in this regard, more modern developments of geometry regard points and lines as undefined terms. model of a modern geometry then consists of specifications of points and lines Definition. n bstract Geometry G consists of a pair {P, L} where P is a set and L is a collection of subsets of P. The elements of P are called Points and the elements of L are called Lines. We will assume that certain statements regarding these points and lines are true at the outset. Statements like these which are assumed true for a geometry are 1
2 called xioms of the geometry. Two xioms we require are that each pair of points P, Q in P belongs to at least one line l in L, and that each line l in L contains at least two elements of P. We can impose further geometric structure by adding other axioms to this definition as the following example of a finite geometry  finite because it contains only finitely many points  illustrates. (Here we have added a third axiom and slightly modified the two mentioned above.) Definition. 4POINT geometry is an abstract geometry = {P, L} in which the following axioms are assumed true: xiom 1: P contains exactly four points; xiom 2: each pair of distinct points in P belongs to exactly one line; xiom 3: each line in L contains exactly two distinct points. The definition doesn t indicate what objects points and lines are in a 4Point geometry, it simply imposes restrictions on them. Only by considering a model of a 4Point geometry can we get an explicit description. Look at a tetrahedron. It has 4 vertices and 6 edges. Each pair of vertices lies on exactly one edge, and each edge contains exactly 2 vertices. Thus we get the following result. Point geometry in which Example. tetrahedron contains a model of a 4 P = {vertices of the tetrahedron} and L = {edges of the tetrahedron}. This example is consistent with our usual thinking of what a point in a geometry should be and what a line should be. But points and lines in a 4Point geometry can be anything so long as they satisfy all the axioms. Exercise provides a very different model of a 4 Point geometry in which the points are opposite faces of an octahedron and the lines are the vertices of the octahedron! 2
3 Why do we bother with models? Well, they give us something concrete to look at or think about when we try to prove theorems about a geometry Theorem. In a 4Point geometry there are exactly 6 lines. To prove this theorem synthetically all we can do is use the axioms and argue logically from those. model helps us determine what the steps in the proof should be. Consider the tetrahedron model of a 4Point geometry. It has 6 edges, and the edges are the lines in the geometry, so the theorem is correct for this model. But there might be a different model of a 4Point geometry in which there are more than 6 lines, or fewer than 6 lines. We have to show that there will be exactly 6 lines whatever the model might be. Let s use the tetrahedron model again to see how to prove this. Label the vertices, B, C, and D. These are the 4 points in the geometry. Concentrate first on. There are 3 edges passing through, one containing B, one containing C, and one containing D; these are obviously distinct edges. This exhibits 3 distinct lines containing. Now concentrate on vertex B. gain there are 3 distinct edges passing through B, but we have already counted the one passing also through. So there are only 2 new lines containing B. Now concentrate on vertex C. Only the edge passing through C and D has not been counted already, so there is only one new line containing C. Finally concentrate on D. Every edge through D has been counted already, so there are no new lines containing D. Since we have looked at all 4 points, there are a total of 6 lines in all. This proof applies to any 4Point geometry if we label the four points, B, C, and D, whatever those points are. xiom 2 says there must be one line containing and B, one containing and C and one containing and D. But the xiom 3 says that the line containing and B must be distinct from the line containing and C, as well as the line containing and D. Thus there will always be 3 distinct lines containing. By the same argument, there will be 3 distinct lines containing B, but one of these will contain, so there are only 2 new lines containing B. Similarly, there will be 1 new line containing C and no new lines containing D. Hence in any 4Point geometry there will be exactly 6 lines. This is usually how we prove theorems in xiomatic Geometry: look at a model, check that the theorem is true for the model, then use the axioms and theorems that follow from 3
4 these axioms to give a logically reasoned proof. For Euclidean plane geometry that model is always the familiar geometry of the plane with the familiar notion of point and line. But it is not be the only model of Euclidean plane geometry we could consider! To illustrate the variety of forms that geometries can take consider the following example Example. Denote by P 2 the geometry in which the points (here called Ppoints) consist of all the Euclidean lines through the origin in 3space and the Plines consist of all Euclidean planes through the origin in 3space. Since exactly one plane can contain two given lines through the origin, there exists exactly one Pline through each pair of Ppoints in P 2 just as in Euclidean plane geometry. But what about parallel Plines? For an abstract geometry G we shall say that two lines m, and l in G are parallel when l and m contain no common points. This makes good sense and is consistent with our usual idea of what parallel means. Since any two planes through the origin in 3space must always intersect in a line in 3space we obtain the following result Theorem. In P 2 there are no parallel Plines. ctually, P 2 is a model of Projective plane geometry. The following figure illustrates some of the basic ideas about P 2. B 4
5 The two Euclidean lines passing through and the origin and through B and the origin specify two Ppoints in P 2, while the indicated portion of the plane containing these lines through and B specify the Pline segment B. Because of Theorem 3.1.6, the geometry P 2 cannot be a model for Euclidean plane geometry, but it comes very close. Fix a plane passing through the origin in 3space and call it the Equatorial Plane by analogy with the plane through the equator on the earth Example. Denote by E 2 the geometry in which the Epoints consist of all lines through the origin in 3space that are not contained in the equatorial plane and the Elines consist of all planes through the origin save for the equatorial plane. In other words, E 2 is what is left of P 2 after one Pline and all the Ppoints on that Pline in P 2 are removed. The claim is that E 2 can be identified with the Euclidean plane. Thus there must be parallel Elines in this new geometry E 2. Do you see why? Furthermore, E 2 satisfies Euclid s Fifth Postulate. The figure below indicates how E 2 can be identified with the Euclidean plane. Look at a fixed sphere in Euclidean 3Space centered at the origin whose equator is the circle of intersection with the fixed equatorial plane. Now look at the plane which is tangent to this sphere at the North Pole of this sphere. 5
6 B Every line through the origin in 3space will intersect this tangent plane in exactly one point unless the line is parallel in the usual 3dimensional Euclidean sense to the tangent plane at the North Pole. But these parallel lines are precisely the lines through the origin that lie in the equatorial plane. On the other hand, for each point in the tangent plane at the North Pole there is exactly one line in 3space passing through both the origin and the given point in the tangent plane. Thus there is a 11 correspondence between the Epoints in E 2 and the points in the tangent plane at the North Pole. In the same way we see that there is a 11 correspondence between Elines in E 2 and the usual Euclidean lines in the tangent plane. The figure above illustrates the 11 correspondence between Eline segment B in E 2 and the line segment B in Euclidean plane geometry. For reasons, which will become very important later in connection with transformations, this 11 correspondence can be made explicit through the use of coordinate geometry and ideas from linear algebra. Let the fixed sphere centered at the origin having radius 1. Then the point (x, y) in the Euclidean plane is identified with the point (x, y, 1) in the tangent plane at the North Pole, and this point is then identified with the line { (x, y, 1): < < } through the origin in 3space. Since there are no parallel lines in P 2 it is clear that the removal from P 2 of that one P line and all Ppoints on that Pline must be very significant Exercise. What points do we need to add to the Euclidean plane so that under the identification of the Euclidean plane with E 2 the Euclidean plane together with these 6
7 additional points are in 11 correspondence with the points in P 2? What line do we need to add to the Euclidean plane so that we get a 11 correspondence with all the lines in P 2? Note first that by restricting further the points and lines in P 2 we get a model of a different geometry. The set of all lines passing through the origin in 3space and through the 45 th parallel in the Northern Hemisphere of the fixed sphere model determines a cone in 3space to be denoted by L. B Definition. Denote by H 2 the geometry whose hpoints consists of Euclidean lines through the origin in 3space that lie in the inside the cone L and whose hlines consist of the intersections of the interior of L and planes through the origin in 3space. gain the Euclidean lines through and B represent hpoints and B in H 2 and the hline segment B is (as indicated in the above figure by the shaded region) the sector of a plane containing the Euclidean lines through the origin which are passing through points on the line segment connecting and B. H 2 is a model of Hyperbolic plane geometry. The reason why it's a model of a 'plane' geometry is clear because we have only defined points and lines, but what is not at all obvious is why the name 'hyperbolic' is used. To understand that let's try to use H 2 to create other models. For instance, our intuition about 'plane' geometries suggests that we should try to find models in which hpoints really are points, 7
8 not lines through the origin! One way of doing this is by looking at surfaces in 3space, which intersect the lines inside the cone L exactly once. There are two natural candidates, both presented here. The second one presented realizes Hyperbolic plane geometry as the points on a hyperboloid,  hence the name 'Hyperbolic' geometry. The first one presented realizes Hyperbolic plane geometry as the points inside a disk. This first one, known as the Klein Model, is very useful for solving the following exercise because its hlines are realized as open Euclidean line segments. In the next section we study a third model known as the Poincaré Disk Exercise. Given an hline l in Hyperbolic plane geometry and an hpoint P not on the hline, how many hlines parallel to l through P are there? Klein Model. Consider the tangent plane M, tangent to the unit sphere at its North Pole, and let the origin in M be the point of tangency of M with the North Pole. Then M intersects the cone L in a circle, call it Σ, and it intersects each line inside L in exactly one point inside Σ. In fact, there is a 11 correspondence between the lines inside L and the points inside Σ. On the other hand, the intersection of M with planes is a Euclidean line, so the lines in H 2 are in 11 correspondence with the chords of Σ, except that we must remember that points on circle Σ correspond to lines on L. So the lines in the Klein model of Hyperbolic plane geometry are exactly the chords of Σ, omitting the endpoints of a chord. In other words, the hyperbolic hlines in this model are open line segments. The following picture contains some points and lines in the Klein model, B D C the dotted line on the circumference indicating that these points are omitted. 8
9 3.1.11a Exercise. Solve Exercise using the Klein model Hyperboloid model. Consider the hyperbola z x = 1 in the x,zplane. Its asymptotes are the lines z = ±x. Now rotate the hyperbola and its asymptotes about the z axis. The asymptotes generate the cone L, and the hyperbola generates a twosheeted hyperboloid lying inside L; denote the upper hyperboloid by B. Then every line through the origin in 3space intersects B exactly once see Exercise ; in fact, there is a 11 correspondence between the points on B and the points in H 2. The lines in H 2 correspond to the curves on B obtained by intersecting the planes though the origin in 3 space. With this model, the hyperboloid B is a realization of Hyperbolic plane geometry Exercise. Prove that every line through the origin in 3space intersects B (in the Hyperbolic model above) exactly once. 3.2 POINCRÉ DISK. lthough the line geometries of the previous section provide a very convenient, coherent, and illuminating way of introducing models of noneuclidean geometries, they are not convenient ones in which to use Sketchpad. More to the point, they are not easy to visualize or to work with. The Klein and Hyperboloid models are more satisfactory ones that conform more closely to our intuition of what a plane geometry should be, but the definition of distance between points and that of angle measure conform less so. We instead focus on the Poincaré Model D, introduced by Henri Poincaré in 1882, where hpoints are points as we usually think of them  points in the plane  while hlines are arcs of particular Euclidean circles. This too fits in with our usual experience of Euclidean plane geometry if one thinks of a straight line through point as the limiting case of a circle through point whose radius approaches as the center moves out along a perpendicular line through. The Poincaré Disk Model allows the use of standard Euclidean geometric ideas in the development of the geometric properties of the models and hence of Hyperbolic plane geometry. We will see later that D is actually a model of the "same" geometry as H 2 by constructing a 11 transformation from H 2 onto D. Let C be a circle in the Euclidean plane. Then D is the geometry in which the hpoints are the points inside C and the hlines are the arcs inside C of any circle intersecting C at right angles. This means that we omit the points of intersection of these circles with C. In addition, any diameter of the bounding circle will also be an hline, since any straight line through the center of the bounding circle intersects the bounding circle at right angles and (as before) can be regarded as the limiting case of a circle whose radius approaches infinity. 9
10 s in the Klein model, points on the circle are omitted and hyperbolic hlines are open   in this case, open arcs of circles. s we are referring to points inside C as hpoints and the hyperbolic lines inside C as hlines; it will also be convenient to call C the bounding circle. The following figure illustrates these definitions: C F G,B,E,F, and G are hpoints. E hsegment hray hline B More technically, we say that a circle intersecting C at right angles is orthogonal to C. Just as for Euclidean geometry, it can be shown that through each pair of hpoints there passes exactly one hline. coordinate geometry proof of this fact is included in Exercise We suggest a synthetic proof of this in Section 3.5. Thus the notion of hline segment between hpoints and B makes good sense: it is the portion between and B of the unique hline through and B. In view of the definition of hlines, the hline segment between and B can also be described as the arc between and B of the unique circle through and B that is orthogonal to C. Similarly, an hray starting at an hpoint in D is either one of the two portions, between and the bounding circle, of an hline passing through. Having defined D, the first two things to do are to introduce the distance, d h (, B), between hpoints and B as well as the angle measure of an angle between hrays starting at some hpoint. The distance function should have the same properties as the usual Euclidean distance, namely: (Positivedefiniteness): For all points and B ( B ), d h (, B) > 0 and d h (, ) = 0; (Symmetry): For all points and B, d h (, B) = d h (B, ); (Triangle inequality): For all points, B and C, d h (, B) d h (,C) + d h (C, B). 10
11 Furthermore the distance function should satisfy the Ruler Postulate Ruler Postulate: The points in each line can be placed in 11 correspondence to the real numbers in such a way that: each point on the line has been assigned a unique real number (its coordinate); each real number is assigned to a unique point on the line; for each pair of points, B on the line, d h (, B) = a  b, where a and b are the respective coordinates of and B. The function we adopt for the distance looks very arbitrary and bizarre at first, but good sense will be made of it later, both from a geometric and transformational point of view. Consider two hpoints, B in D and let M, N be the points of intersection with the bounding circle of the hline through, B as in the figure: C N B M We set d h d(, M)d(B, N) (, B) = ln d(, N)d(B,M) where d(, M) is the usual Euclidean distance between points and M. Using properties of logarithms, one can check that the role of M and N can be reversed in the above formula (see Exercise 3.3.7) Exercise. Show that d h (, B) satisfies the positivedefiniteness and symmetry conditions above. 11
12 We now introduce angles and angle measure in D. Just as in the Euclidean plane, two hrays starting at the same point form an angle. In the figure below we see two intersecting hlines forming BC. θ B C To find the hyperbolic measure m h BC of BC we appeal to angle measure in Euclidean geometry. To do that we need the tangents to the arcs at the point. The hyperbolic measure of the angle BC is then defined to be the Euclidean measure of the Euclidean angle between these two tangents, i.e. m h BC = m. Just as the notions of points, lines, distance and angle measure are defined in Euclidean plane geometry, these notions are all defined in D. nd, we can exploit the hyperbolic tools for Sketchpad, which correspond to the standard Euclidean tools, to discover facts and theorems about the Poincaré Disk and hyperbolic plane geometry in general. Load the Poincaré folder of scripts by moving the sketch Poincare Disk.gsp into the Tool Folder. To access this sketch, first open the folder Samples, then Sketches, then Investigations. Once Sketchpad has been restarted, the following scripts will be available: Hyperbolic Segment  Given two points, constructs the hsegment joining them Hyperbolic Line  Constructs an hline through two hpoints Hyperbolic P. Bisector  Constructs the perpendicular bisector between two h points Hyperbolic Perpendicular  Constructs the perpendicular of an hline through a third point not on the hline. Hyperbolic. Bisector Constructs an hangle bisector. Hyperbolic Circle By CP  Constructs an hcircle by center and point. 12
13 Hyperbolic Circle By CR Constructs an hcircle by center and radius. Hyperbolic ngle Gives the hyperbolic angle measure of an hangle. Hyperbolic Distance  Gives the measure of the hyperbolic distance between two hpoints which do not both lie on a diameter of the Poincare disk. The sketch Poincare Disk.gsp contains a circle with a specially labeled center called, P. Disk Center, and point on the disk called, P. Disk Radius. The tools listed above work by using utomatching to these two labels, so if you use these tools in another sketch, you must either label the center and radius of your Poincare Disk accordingly, or match the disk center and radius before matching the other givens for the tool. We are now ready to investigate properties of the Poincaré Disk. Use the line tool to investigate how the curvature of hlines changes as the line moves from one passing close to the center of the Poincaré disk to one lying close to the bounding circle. Notice that this line tool never produces hlines passing through the center of the bounding circle for reasons that will be brought out in the next section. In fact, if you experiment with the tools, you will find that the center of the Poincare Disk and the hlines which pass through the center are problematic in general. Special tools need to be created to deal with these cases. (There is another very good software simulation of the Poincaré disk available on the web at You can download the program or run it online. The site also contains some background material that you may find interesting.) Demonstration: Parallel Lines. s in Euclidean geometry, two hlines in D are said to be parallel when they have no hpoints in common. In the Poincaré disk construct an hline l and an hpoint P not on l. Use the hline script to investigate if an hline through P parallel to l can be drawn. Can more than one be drawn? How many can be drawn? End of Demonstration Shortest Distance. In Euclidean plane geometry the line segment joining points P and Q is the path of shortest distance; in other words, a line segment can be described both in metric terms and in geometric terms. More precisely, there are two natural definitions of 13
14 a line segment PQ, one as the shortest path between P and Q, a metric property, the other as all points between P, Q on the unique line l passing through P and Q  a geometric property. But what do we mean by between? That is easy to answer in terms of the metric: the line segment PQ consists of all points R on l such that d h (P, R) + d h (R, Q) = d h (P, Q). This last definition makes good sense also in D since there we have defined a notion of distance a Demonstration: Shortest Distance. In the Poincaré disk select two points and B. Use the Hyperbolic Distance tool to investigate which points C minimize the sum d h (C, ) + d h (C, B). What does your answer say about an hline segment between and B? End of Demonstration 3.2.3a Demonstration: Hyperbolic Versus Euclidean Distance. Since Sketchpad can measure both Euclidean and hyperbolic distances we can investigate hyperbolic distance and compare it with Euclidean distance. Draw two hline segments, one near the center of the Poincaré disk, the other near the boundary. djust the segments until both have the same hyperbolic length. What do you notice about the Euclidean lengths of these arcs? Compute the ratio d h (, B) d(, B) of the hyperbolic and Euclidean lengths of the respective hyperbolic and Euclidean line segments between points, B in the Poincaré disk. What is the largest value you can obtain? End of Demonstration Demonstration: Investigating d h further. Does this definition of d h depend on where the boundary circle lies in the plane? What is the effect on d h if we change the center of the circle? What is the effect on d h of doubling the radius of the circle? By changing the size of the disk, but keeping the points in the same proportion we can answer these questions. Draw an hline segment B and measure its length. Over on the toolbar change the select arrow to the Dilate tool. Select P. Disk Center, then Transform Mark Center. Under the Edit menu Select ll, then 14
15 deselect the Distance =. Now, without deselecting these objects, drag the P. Disk Radius to vary the size of the PDisk and of all the Euclidean distances between objects inside proportionally. What effect does changing the size of the PDisk proportionally (relative to the PDisk Center) have on the hyperbolic distance between the two endpoints of the hyperbolic segment? Over on the toolbar change the select arrow to the Rotate tool. Select PDisk Center, then Transform Mark Center. Under the Edit menu Select ll, then deselect the Distance =. Now, without deselecting these objects, drag the PDisk Radius to rotate the orientation of the PDisk. What effect does changing the orientation of the PDisk uniformly have on the hyperbolic distance between the two endpoints of the hyperbolic segment? Over on the toolbar, change the Rotate tool back to the select arrow. Under the Edit menu Select ll, then deselect the Distance =. Grab the PDisk Center, and drag the Disk around the screen. What effect does changing the location of the PDisk have on the hyperbolic distance between the two endpoints of the hline segment? End of Demonstration Exercises. This Exercise set contains questions related to bstract Geometries and properties of the Poincaré Disk. Exercise Prove that in a 4Point geometry there passes exactly 3 lines through each point. Exercise The figure to the right is an octahedron. Use this to exhibit a model of a 4Point geometry that is very different from the tetrahedron model we used in class. Four of the faces have been picked out. Use these as the 4 points. What must the lines be if the octahedron is to be a model of a 4 Point geometry? Make sure you check that all the axioms of a 4Point geometry are satisfied. Exercise We have stated that our definition for the hyperbolic distance between two points satisfies the ruler postulate, but it is not easy to construct very long hline segments, say ones of length 10. The source of this difficulty is the rapid growth of the exponential 15
16 function. Suppose that the radius of the bounding circle is 1 and let be an hpoint that has Euclidean distance r from the origin (r < 1, of course). The diameter of the bounding circle passing through is an hline. Show the hyperbolic distance from the center of the bounding circle to is (1+ r) ln (1 r) ; Find r when the hyperbolic distance from to the center of the bounding circle is 10. Exercise Use Exercise to prove that the second statement of the ruler postulate holds when the hyperbolic line is a diameter of the bounding circle and if to each point we assign the hyperbolic distance between it and the center of the bounding circle. That is, why are we guaranteed that each real number is assigned to a unique point on the line? Hint: Show your function for r from Exercise is 11 and onto the interval (1,1). Exercise Explain why the ruler postulate disallows the use of the Euclidean distance formula to compute the distance between two points in the Poincaré Disk. Exercise Using Sketchpad open the Poincaré Disk Starter and find a counterexample within the Poincaré Disk to each of the following. (a) If a line intersects one of two parallel lines, then it intersects the other. (b) If two lines are parallel to a third line then the two lines are parallel to each other. Exercise Using properties of logarithms and properties of absolute value, show that, with the definition of hyperbolic distance, d h d(, M)d(B, N) (, B) = ln d(, N)d(B,M) d(, N)d(B, M) = ln, d(, M)d(B, N) i.e., the roles of M and N can be reversed and the same distance value results. 3.4 CLSSIFYING THEOREMS. For many years mathematicians attempted to deduce Euclid's fifth postulate from the first four postulates and five common notions. Progress came in the nineteenth century when mathematicians abandoned the effort to find a contradiction in the denial of the fifth postulate and instead worked out carefully and completely the consequences of such a denial. It was found that a coherent theory arises if one assumes the Hyperbolic Parallel Postulate instead of Euclid's fifth Postulate. 16
17 Hyperbolic Parallel Postulate: Through a point P not on a given line l there exists at least two lines parallel to l. The axioms for hyperbolic plane geometry are Euclid s 5 common notions, the first four postulates and the Hyperbolic Parallel Postulate. Three professional mathematicians are credited with the discovery of Hyperbolic geometry. They were Carl Friedrich Gauss ( ), Nikolai Ivanovich Lobachevskii ( ) and Johann Bolyai ( ). ll three developed noneuclidean geometry axiomatically or on a synthetic basis. They had neither an analytic understanding nor an analytic model of noneuclidean geometry. Fortunately, we have a model now; the Poincaré disk D is a model of hyperbolic plane geometry, meaning that the five axioms, consisting of Euclid s first four postulates and the Hyperbolic Parallel Postulate, are true statements about D, and so any theorem that we deduce from these axioms must hold true for D. In particular, there are several lines though a given point parallel to a given line not containing that point. Now, an abstract geometry (in fact, any axiomatic system) is said to be categorical if any two models of the system are equivalent. When a geometry is categorical, any statement which is true about one model of the geometry is true about all models of the geometry and will be true about the abstract geometry itself. Euclidean geometry and the geometries that result from replacing Euclid s fifth postulate with lternative or lternative B are both categorical geometries. In particular, Hyperbolic plane geometry is categorical and the Poincaré disk D is a model of hyperbolic plane geometry. So any theorem valid in D must be true of Hyperbolic plane geometry. To prove theorems about Hyperbolic plane geometry one can either deduce them from the axioms (i.e., give a synthetic proof) or prove them from the model D (i.e., give an analytic proof). Since both the model D and Hyperbolic plane geometry satisfy Euclid s first four postulates, any theorems for Euclidean plane geometry that do not require the fifth postulate will also be true for hyperbolic geometry. For example, we noted in Section 1.5 that the proof that the angle bisectors of a triangle are concurrent is independent of the fifth postulate. By comparison, any theorem in Euclidean plane geometry whose proof used the Euclidean fifth postulate might not be valid in hyperbolic geometry, though it is not automatically ruled out, as there may be a proof that does not use the fifth postulate. For example, the proof we gave of the existence of the centroid used the fifth postulate, but other proofs, independent of the fifth postulate, do exist. On the other hand, all proofs of the existence of the circumcenter must rely in some way on the fifth postulate, as this result is false in hyperbolic geometry. 17
18 Exercise fter the proof of Theorem 1.5.5, which proves the existence of the circumcenter of a triangle in Euclidean geometry, you were asked to find where the fifth postulate was used in the proof. To answer this question, open a sketch containing a Poincare Disk with the center and radius appropriately labeled (P. Disk Center and P. Disk Radius). Draw a hyperbolic triangle and construct the perpendicular bisectors of two of the sides. Drag the vertices of the triangle and see what happens. Do the perpendicular bisectors always intersect? Now review the proof of Theorem and identify where the Parallel postulate was needed. We could spend a whole semester developing hyperbolic geometry axiomatically! Our approach in this chapter is going to be either analytic or visual, however, and in chapter 5 we will begin to develop some transformation techniques once the idea of Inversion has been adequately studied. For the remainder of this section, therefore, various objects in the Poincaré disk D will be studied and compared to their Euclidean counterparts Demonstration: Circles. circle is the set of points equidistant from a given point (the center). Open a Poincaré Disk, construct two points, and label them by and O. Measure the hyperbolic distance between and O, d h (, O), Select the point and under the Display menu select Trace Points. Now drag while keeping d h (, O) constant. Can you describe what a hyperbolic circle in the Poincaré Disk should look like? To confirm your results, use the circle script to investigate hyperbolic circles in the Poincaré Disk. What do you notice about the center? End of Demonstration Demonstration: Triangles. triangle is a threesided polygon; two hyperbolic triangles are said to be congruent when they have congruent sides and congruent interior angles. Investigate hyperbolic triangles in the Poincaré Disk. Construct a hyperbolic triangle BC and use the Hyperbolic ngle tool to measure the hyperbolic angles of BC (keep in mind that three points are necessary to name the angle, the vertex should be the second point clicked). Calculate the sum of the three angle measures. Drag the vertices of the triangle around. What is a lower bound for the sum of the hyperbolic angles of a triangle? What is an 18
19 upper bound for the sum of the hyperbolic angles of a triangle? What is an appropriate conclusion about hyperbolic triangles? How does the sum of the angles change as the triangle is dragged around D? The proofs of SSS, SS, S, and HL as valid shortcuts for showing congruent triangles did not require the use of Euclid s Fifth postulate. Thus they are all valid shortcuts for showing triangles are congruent in hyperbolic plane geometry. Use SSS to produce two congruent hyperbolic triangles in D. Drag one triangle near the boundary and one triangle near the center of D. What happens? We also had, SSS, and SS shortcuts for similarity in Euclidean plane geometry. Is it possible to find two hyperbolic triangles that are similar but not congruent? Your answer should convince you that it is impossible to magnify or shrink a triangle without distortion! End of Demonstration Demonstration: Special Triangles. n equilateral triangle is a triangle with 3 sides of equal length. n isosceles triangle has two sides of equal length. Create a tool that constructs hyperbolic equilateral triangles in the Poincaré disk. Is an equilateral triangle equiangular? re the angles always 60 as in Euclidean plane geometry? Can you construct a hyperbolic isosceles triangle? re angles opposite the congruent sides congruent? Does the ray bisecting the angle included by the congruent sides bisect the side opposite? Is it also perpendicular? How do your results compare to Theorem and Corollary 1.4.7? End of Demonstration Demonstration: Polygons. rectangle is a quadrilateral with four right angles. Is it possible to construct a rectangle in D? regular polygon has congruent sides and congruent interior angles. To construct a regular quadrilateral in the Poincaré Disk start by constructing an hcircle and any diameter of the circle. Label the intersection points of the diameter and the circle as and C. Next construct the perpendicular bisector of the diameter and label the intersection points with the circle as B and D. Construct the line segments B, BC, CD, and D,. 19
20 D C B BCD is a regular quadrilateral. Then BCD is a regular quadrilateral. Why does this work? Create a tool from your sketch. The following theorems are true for hyperbolic plane geometry as well as Euclidean plane geometry: ny regular polygon can be inscribed in a circle. ny regular polygon can be circumscribed about a circle. Consequently, any regular ngon can be divided into n congruent isosceles triangles just as in Euclidean plane geometry. Modify the construction to produce a regular octagon and regular 12gon. Create tools from your sketches. End of Demonstration By now you may have started to wonder how one could define area within hyperbolic geometry. In Euclidean plane geometry there are two natural ways of doing this, one geometric, the other analytic. In the geometric definition we begin with the area of a fixed shape, a square, and then build up the area of more complicated figures as sums of squares so that we could say that the area of a figure is n square inches, say. Since squares don t exist in hyperbolic plane geometry, however, we cannot proceed in this way. Now any definition of area should have the following properties: Every polygonal region has one and only one area, (a positive real number). Congruent triangles have equal area. If a polygonal region is partitioned into a pair of sub regions, the area of the region will equal the sum of the areas of the two sub regions. Recall, that in hyperbolic geometry we found that the sum of the measures of the angles of any triangle is less than 180. Thus we will define the defect of a triangle as the amount by which the angle sum of a triangle misses the value Definition. The defect of triangle BC is the number ( BC) =180 m m B m C 20
21 More generally, the defect can be defined for polygons Definition. The defect of polygon P 1 P 2 KP n is the number ( P 1 P2 KPn ) = 180( n 2) m P1 m P2 K m P n It may perhaps be surprising, but this will allow us to define a perfectly legitimate area function where the area of a polygon P 1 P 2 KP n is k times its defect. The value of k can be specified once a unit for angle measure is agreed upon. For example if our unit of angular measurement is degrees, and we wish to express angles in terms of radians then we use the constant k = 180 o. It can be shown that this area function defined below will satisfy all of the desired properties listed above Definition. The area rea h (P 1 P 2 KP) of a polygon P 1 P 2 KP n is defined by where k is a positive constant. rea h (P 1 P 2 KP n ) = k (P 1 P 2 KP n ) Note, that this puts an upper bound on the area of all triangles, namely 180 k. (More generally, 180 (n 2) k for ngons.) This definition becomes even stranger when we look at particular examples a Demonstration: reas of Triangles. Open a Poincare disk. Construct a hyperbolic 'triangle' h OMN having one vertex O at the origin and the remaining two vertices M, N on the bounding circle. This is not a triangle in the strict sense because points on the bounding circle are not points in the Poincare disk. Nonetheless, it is the limit of a hyperbolic triangle h OB as, B approach the bounding circle. 21
22 Disk N O B M PDisk Radius The 'triangle' h OMN is called a Doublysymptotic triangle. Determine the length of the hyperbolic line segment B using the length script. Then measure each of the interior angles of the triangle and compute the area of h OB (use k=1). What happens to these values as, B approach M, N along the hyperbolic line through, B? Set rea h ( h OMN) = lim rea h ( h OB) Explain this value by relating it to properties of h OMN. Repeat this construction, replacing the center O by any point C in the Poincaré disk. 22
23 Disk C N PDisk Center B M PDisk Radius What value do you obtain for rea h ( h CB)? Now let, B approach M, N along the hyperbolic line through, B and set rea h ( h CMN) = lim rea h ( h CB); again we say that h (CMN) is a doublyasymptotic triangle. Relate the value of rea h ( h CMN) to properties of h CMN. Select an arbitrary point L on the bounding circle and let C approach L. We call h LMN a triplyasymptotic triangle. Now set rea h ( h LMN) = lim rea h ( h CMN). Explain your value for rea h ( h LMN) in terms of the properties of h LMN. End of Demonstration 3.4.7a. Your investigations may lead you to conjecture the following result. 23
24 3.4.8 Theorem. (a) The area of a hyperbolic triangle is at most 180k even though the lengths of its sides can be arbitrarily large. (b) The area of a triplyasymptotic triangle is always 180k irrespective of the location of its vertices on the bounding circle. By contrast, in Euclidean geometry the area of a triangle can become unboundedly large as the lengths of its sides become arbitrarily large. In fact, it can be shown that Euclid s Fifth Postulate is equivalent to the statement: there is no upper bound for the areas of triangles Summary. The following results are true in both Euclidean and Hyperbolic geometries: SS, S, SSS, HL congruence conditions for triangles. Isosceles triangle theorem (Theorem and Corollary 1.4.7) ny regular polygon can be inscribed in a circle. The following results are strictly Euclidean Sum of the interior angles of a triangle is180. Rectangles exist. The following results are strictly Hyperbolic The sum of the interior angles of a triangle is less than 180. Parallel lines are not everywhere equidistant. ny two similar triangles are congruent. Further entries to this list are discussed in Exercise set 3.6. s calculus showed, there is also an analytic way introducing the area of a set in the Euclidean plane as a double integral dxdy. 24
25 n entirely analogous analytic definition can be made for the Poincaré disk. What is needed is a substitute for dxdy. If we use standard polar coordinates ( r, ) for the Poincaré disk, then the hyperbolic area of a set is defined by rea h ( ) = 4rdrd 1 r 2. Of course, when is an ngon, it has to be shown that this integral definition of area coincides with the value defined by the defect of up to a fixed constant k independent of. Calculating areas with this integral formula often requires a high degree of algebraic ingenuity, however. 3.5 ORTHOGONL CIRCLES. Orthogonal circles, i.e. circles intersecting at right angles, arise on many different occasions in plane geometry including the Poincaré disk model D of hyperbolic plane geometry introduced in the previous section. In fact, their study constitutes a very important part of Euclidean plane geometry known as Inversion Theory. This will be studied in some detail in Chapter 5, but here we shall develop enough of the underlying ideas to be able to explain exactly how the tools constructing hlines and hsegments are obtained. Note first that two circles intersect at right angles when the tangents to both circles at their point of intersection are perpendicular. nother way of expressing this is say that the tangent to one of the circles at their point of intersection D passes through the center of the other circle as in the figure below. D O Does this suggest how orthogonal circles might be constructed? 25
26 3.5.1 Exercise. Given a circle C 1 centered at O and a point D on this circle, construct a circle C 2 intersecting C 1 orthogonally at D. How many such circles C 2 can be drawn? It should be easily seen that there are many possibilities for circle C 2. By requiring extra properties of C 2 there will be only one possible choice of C 2. In this way we see how to construct the unique hline through two points P, Q in D Demonstration. Given a circle C 1 centered at O, a point not on C 1, as well as a point D on C 1, construct a circle C 2 passing through and intersecting C 1 at D orthogonally. How many such circles C 2 can be drawn? C 1 D O Sketchpad provides a very illuminating solution to this problem. Open a new sketch. Draw circle C 1, labeling its center O, and construct point not on the circle as well as a point D on the circle. Construct the tangent line to the circle C 1 at D and then the segment D. Construct the perpendicular bisector of D. The intersection of this perpendicular bisector with the tangent line to the circle at D will be the center of a circle passing through both and D and intersecting the circle C 1 orthogonally at D. Why? The figure below illustrates the construction when is outside circle C 1. 26
27 C 1 D O C 2 What turns out to be of critical importance is the locus of circle C 2 passing through and D and intersecting the given circle C 1 orthogonally at D, as D moves. Use Sketchpad to explore the locus. Select the circle C 2, and under the Display menu select trace circle. Drag D. lternatively you can select the circle C 2, then select the point D and under the Construct menu select locus. The following figure was obtained by choosing different D on the circle C 1 and using a script to construct the circle through (outside C 1 ) and D orthogonal to C 2. The figure you obtain should look similar to this one, but perhaps more cluttered if you have traced the circle. 27
28 C 1 O B D C 2 Your figure should suggest that all the circles orthogonal to the given circle C 1 that pass through have a second common point on the line through O (the center of C 1 ) and. In the figure above this second common point is labeled by B. [Does the figure remind you of anything in Physics  the lines of magnetic force in which the points and B are the poles of the magnet. say?] Repeat the previous construction when is inside C 1 and you should see the same result. End of Demonstration t this moment, Theorem and its converse will come into play Theorem. Fix a circle C 1 with center O, a point not on the circle, and point D on the circle, Now let B be the point of intersection of the line through O with the circle through and D that is orthogonal to C 1. Then B satisfies O.OB = OD 2. In particular, the point B is independent of the choice of point D. The figure below illustrates the theorem when is outside C 1. 28
29 C 1 O B D O = 1.65 inches OB = 0.52 inches O OB = 0.86 inches 2 OD 2 = 0.86 inches 2 Proof. By construction the segment OD is tangential to the orthogonal circle. Hence O.OB = OD 2 by Theorem QED Theorem has an important converse Theorem. Let C 1 be a circle of radius r centered at O. Let and B be points on a line through O (neither or B on C 1 ). If O.OB = r 2 then any circle through and B will intersect the circle C 1 orthogonally. Proof. Let D denote a point of intersection of the circle C 1 with any circle passing through and B. Then O.OB = OD 2. So by Theorem 2.9.4, the line segment OD will be tangential to the circle passing through, B, and D. Thus the circle centered at O will be orthogonal to the circle passing through, B, and D. QED Theorems and can be used to construct a circle orthogonal to a given circle C 1 and passing through two given points P, Q inside C 1. In other words, we can show how to construct the unique hline through two given points P, Q in the hyperbolic plane D Demonstration. Open a new sketch and draw the circle C 1, labeling its center by O. Now select arbitrary points P and Q inside C 1. Choose any point D on C 1. Construct the circle C 2 passing through P and D that is orthogonal to C 1. Draw the ray starting at O and passing through P. Let B be the other point of intersection of this ray with C 2. By Theorem OP.OB = OD 2. Confirm this by measuring OP, OB, and OD in your figure. 29
30 Construct the circumcircle passing through the vertices P, Q and B of PQB. By Theorem this circumcircle will be orthogonal to the given circle. OP OB = 1.15 inches 2 OD 2 = 1.15 inches 2 C 1 O Q P D C 2 B If P and Q lie on a diameter of C 1 then the construction described above will fail. Why? This explains why there had to be separate scripts in Sketchpad for constructing hlines passing through the center of the bounding circle of the Poincaré model D of hyperbolic plane geometry. The points, B described in Theorem are said to be Inverse Points. The mapping taking to B is said to be Inversion. The properties of inversion will be studied in detail in Chapter 5 in connection with tilings of the Poincaré model D. Before then in Chapter 4, we will study transformations. End of Demonstration Exercises. In this set of exercises, we ll look at orthogonal circles, as well as other results about the Poincaré Disk. Exercise To link up with what we are doing in class on orthogonal circles, recall first that the equation of a circle C in the Euclidean plane with radius r and center (h, k) is (x  h) 2 + (y  k) 2 = r 2 which on expanding becomes x 22hx + y 22ky + h 2 + k 2 = r 2. Now consider the special case when C has center at the origin (0, 0) and radius 1. Show that the equation of the circle orthogonal to C and having center (h, k) is given by x 22hx + y 22ky + 1 = 0. Exercise One very important use of the previous problem occurs when C is the bounding circle of the Poincaré disk. Let = (a 1, a 2 ) and B = (b 1, b 2 ) be two points inside the circle C, i.e., two hpoints. Show that there is one and only one choice of (h, k) for which 30
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